Reduced and Irreducible Simple Algebraic Extensions of Commutative Rings

Mathematica Moravica Vol. 21, No. 1 (2017), 69–86

Reduced and irreducible simple algebraic extensions of commutative rings

S.V. Mihovski

Abstract. Let A be a commutative ring with identity and α be an algebraic element over A. We give necessary and suﬃcient conditions under which the simple algebraic extension A[α] is without nilpotent or without idempotent elements.

1. Introduction Let A be a commutative ring with identity element 1. We shall say that K is a commutative ring extension of A, or A is a subring of K, if A and K are commutative rings with common identity element and A ⊆ K. Suppose that K is a commutative ring extension of A and let α ∈ K be an algebraic element overA. If α is a root of a nonzero polynomial f(x) ∈ A[x] of a minimal degree n, then we shall say that f(x) is a minimal polynomial of α over the ring A. The intersection of all subrings of K, containing A and α, we shall denoted by A[α]. The ring A[α] is called a simple algebraic extension of A, which is obtained by adjoining α to A. Let f(x) be any nonzero polynomial over the ring A. If the leading coef- ﬁcient of f(x) is a0 = 1, then f(x) is said to be a monic polynomial over A. And what is more, if the leading coeﬃcient a0 of f(x) is a regular element in A, i.e. a0 is not zero divisor in A, then we shall say that f(x) is a regular polynomial over the ring A. Recall that the ring A is called a reduced ring if A has no nonzero nilpotent elements. The ring A is said to be irreducible if A has no nontrivial idempotents. A main result in [11] asserts that if A is a reduced commutative ring, f(x) is a monic minimal polynomial of α over A and the discriminant ∆(f) is a regular element in A, then the simple algebraic extension A[α] is a reduced ring. Also in [11] is proved that if A is an irreducible ring and the minimal polynomial f(x) of α is monic and irreducible over A, then the simple algebraic extension A[α] again is irreducible. So here arises the

2010 Mathematics Subject Classiﬁcation. 13B25, 13B02. Key words and phrases. Commutative rings, Polynomials, Discriminants, Resultants, Simple algebraic extensions, Reduced rings, Irreducible rings.

c 2017 Mathematica Moravica 69 70 Reduced and irreducible simple algebraic extensions. . . problem to ﬁnd necessary and suﬃcient conditions under which the ring A[α] is reduced or irreducible. In this paper we solve these two problems in the parts 3 and 4, respectively.

2. Preliminary lemmas and definitions It is well known that every polynomial f(x) of degree n with coefficients from a field F has at most n roots in every field extension of F . Moreover, there exists a field extension F ⊇ F such that F contains exactly n roots of f(x). But this fact does not hold for the ring extensions of F . For example, let G be a direct product of m ≥ 2 cyclic groups of order n and let K = FG be the group ring of the group G over the field F . Then K is a ring extension of F and every element of G is a root of the polynomial f(x) = xn −1 ∈ F [x]. Thus f(x) has at least nm roots in K. Let n n−1 (1) f (x) = a0x + a1x + ··· + an−1x + an (n ≥ 1) be a polynomial over A of degree n. We shall say that the elements α1, α2, . . . , αn form a canonical system roots of the polynomial f(x) ∈ A[x] if there exists a commutative ring extension K ⊇ A such that α1, α2, . . . , αn ∈ K and

(2) f(x) = a0(x − α1)(x − α2) ··· (x − αn). Every regular polynomial over A has at last one canonical system roots [9]. But example shows that over some rings A there exist polynomials that have not roots. Moreover, there exist polynomials which have roots, but they have not canonical systems of roots. For more details see [9]. Later on we shall use the following two deﬁnitions. A discriminant of the polynomial (1) we shall call the following determinant of order 2n − 1

1 a1 ... an 0 0 ... 0 0 a0 a1 ... an 0 ... 0 ...... 0 0 ... 0 a a ... a ∆(f) = ε 0 1 n , n (n−1)a1 (n−2)a2 ··· an−1 0 ... 0 0 na0 (n−1)a1 ... 2an−2 an−1 ... 0 ...... 0 0 ... 0 na0 (n−1)a1 ... an−1

n(n−1) where ε = (−1) 2 . So, for n = 1 and n = 2 we have ∆(f) = 1 and 2 ∆(f) = a1 − 4a0a2, respectively. Let

m m−1 (3) g(x) = b0x + b1x + ··· + bm−1x + bm (m ≥ 1) be another polynomial over A. Then the determinant of order n + m S.V. Mihovski 71

a0 a1 ... an 0 ... 0 0 0 a0 a1 ... an 0 ... 0 ...... R(f, g) = 0 0 ... a0 a1 ... an−1 an b0 b1 ... bm 0 ... 0 0 0 b0 b1 ... bm 0 ... 0 ...... 0 0 ... b0 b1 ... bm−1 bm is said to be a resultant of the polynomials f(x) and g(x) (see [3, 8, 13]). If e1, e2, . . . , ek is a full orthogonal system idempotents of the ring A, that is the ring A is a direct sum of the ideals eiA (i = 1, . . . , k), then for the polynomials f(x), g(x) ∈ A[x] we put fi(x) = eif(x) and gi(x) = eig(x). So from the deﬁnition of R(f, g) we conclude that

R(f, g) = R(f1, g1) + R(f2, g2) + ··· + R(fk, gk). Likewise, ∆(f) = ∆(f1) + ∆(f2) + ··· + ∆(fk). Later on we shall use these facts without special stipulations. Let α1, α2, . . . , αn be a canonical system roots of the polynomial (1). In [9] it is proved that m (4) R(f, g) = a0 g(α1)g(α2) ··· g(αn) for every polynomial g(x) ∈ A[x], even when g(x) does not have roots. Moreover, if n ≥ 2 and f 0(x) is the prime derivative of f(x), then [10]

n(n−1) n−2 0 0 0 (5) ∆(f) = (−1) 2 · a0 · f (α1) · f (α2) ··· f (αn). When A is a ﬁeld, then (4) and (5) show the well known facts that R(f, g) = 0 if and only if f(x) and g(x) have common roots and f(x) has multiple roots if and only if ∆(f) = 0. For arbitrary polynomials f(x), g(x) ∈ A [x] we have the following. Lemma 1. (see [9], also [8] p.159 and [13] p. 130) Let A be any commutative ring with identity. If (1) and (3) are polynomials over A, then (i) There exist polynomials ϕ(x), ψ(x) ∈ A[x] such that deg ϕ(x) ≤ m − 1, deg ψ(x) ≤ n − 1 and R(f, g) = ϕ(x)f(x) + ψ(x)g(x). (ii) If n ≥ 2, then there exist polynomials u(x), v(x) ∈ A [x] such that deg u(x) ≤ n − 2, deg v(x) ≤ n − 1 and ∆(f) = u(x)f(x) + v(x)f 0(x). So we obtain Corollary 1. [9] Let f(x) and g(x) be polynomials over the ring A. (i) If f(x) and g(x) have common roots, then R(f, g) = 0. (ii) If f(x) has multiple roots, then ∆(f) = 0. 72 Reduced and irreducible simple algebraic extensions. . .

Examples show that the converse statements of the preceding corollary in the general case are erroneous [9]. Let S be a multiplicatively closed set of regular elements in A, that is S contains the product of every his two elements and every element of S is not zero divisor in A. Then there exists a ring of quotients S−1A with respect to S ([12], p. 146). Every element of S−1A is of the form s−1a, where s ∈ S, a ∈ A. Thus all elements of S are invertible in S−1A. If S is the set of all regular elements in A, then S−1A is said to be the classical ring of quotients, that we shall denote by Q(A). Lemma 2. Let A be a ring with identity and f(x) be a regular polynomial over A of degree n ≥ 1. Then there exists a ring extension K ⊇ A such that f(x) is a minimal polynomial over A of some element α ∈ K.

Proof. Let (1) be a regular polynomial over A. First, suppose that a0 = 1. If deg f(x) = n = 1, then we put α = −a1 and the statements are trivial. If n ≥ 2, then let K = A[y]/I be the quotient ring of the polynomial ring A[y] modulo the principal ideal I, generated by the polynomial f(y). Thus A¯ = {a + I | a ∈ A} is a subring of K and, because a0 = 1, it is clear that A and A are isomorphic rings. So A can be viewed as a subring of K. Obviously, f(x) is a minimal polynomial of the element α = y + I ∈ K. If a0 6= 1, then let P = Q(A) be the classical ring of quotients of A. Now A ⊆ P , f(x) ∈ P [x] and a0 is an invertible element of P . Therefore −1 g(x) = a0 f(x) is a monic polynomial over P and the statement for g(x) holds. Thus we conclude that there exists a commutative ring extension K ⊇ P such that g(x) is a minimal polynomial over P of some element α ∈ K. Then it is clear that f(x) = a0g(x) is a minimal polynomial of α over A, as was to be shoved. When α ∈ K and P = Q(A) ⊆ K, then by definition there exists the simple ring extensions P [α] with A[α] ⊆ P [α]. But, if P 6⊂ K, then P [α] is not defined in the general case. Later on we shall use the following Lemma 3. Let K ⊇ A be a ring extension and α ∈ K be an algebraic element over A with a regular minimal polynomial f(x) ∈ A[x]. Then there exists a ring extension K1 ⊇ A such that Q(A) ⊆ K1, f(x) is a minimal ∼ polynomial of some β ∈ K1 and A[α] = A[β]. Proof. Let (1) be a minimal polynomial over A of the element α. If P = Q(A) ⊆ K, then we put K1 = K and β = α. Suppose that P 6⊂ K. Since f(x) ∈ P [x], by Lemma 2 it follows that there exists a commutative ring extension K1 ⊇ P such that f(x) is a minimal polynomial over P of some element β ∈ K1. Since A ⊆ P and f(x) ∈ A[x], it is clear that f(x) is a minimal polynomial of β and over A. Now we shall prove that A[α] and A[β] are isomorphic. First we shall show that for g(x) ∈ A[x] the conditions g(α) = 0 and g(β) = 0 are equivalent. Really, since the leading coefficient a0 is invertible S.V. Mihovski 73 in P = Q(A), we have g(x) = f(x)q(x) + r(x), deg r(x) < deg f(x), where the polynomials q(x) and r(x) are with coefficients in P . It is clear k k that there exists a power a0 (k ≥ 1) of the element a0 ∈ A such that a0q(x) k and a0r(x) to be elements of A[x]. Then k k k k a0g(x) = f(x)[a0q(x)] + a0r(x), deg(a0r(x)) < deg f(x). k If g(α) = 0, then a0r(α) = 0 and by the minimum condition of f(x) we k k k conclude that a0g(x) = f(x)[a0q(x)]. Therefore a0g(β) = 0 and so g(β) = 0, k because a0 is an invertible element of P . In similar way from g(β) = 0 we receive g(α) = 0. Now it is easy to verify that the map g(α) 7→ g(β) for all g(x) ∈ A[x] is an isomorphism between A[α] and A[β], as was to be showed. Lemma 4. Let P = Q(A) be the classical ring of quotients of a commutative reduced ring A and let f(x) ∈ A[x] be a regular minimal polynomial of the algebraic element α. Then (i) The rings A[α] and A = A[x]/(A[x] T f(x)P [x]) are isomorphic. (ii) The ring A[α] is reduced if and only if the quotient ring P = P [x]/f(x)P [x] is reduced. Proof. In view the preceding lemma we may assume that Q(A) ⊆ K and α ∈ K. (i) The mapping Φ: A[x] → A[α], defined by Φ(g(x)) = g(α) for all g(x) ∈ A[x], is a homomorphism of A[x] onto A[α] with ker Φ = A[x] T f(x)P [x]. Really, it is clear that A[x] T f(x)P [x] ⊆ ker Φ. If g(x) ∈ ker Φ, then g(α) = f(α) = 0. Moreover, there exist polynomials q(x), r(x) ∈ P [x], such that g(x) = f(x)q(x) + r(x) and deg r(x) < deg f(x). Hence it follows that r(α) = 0. Since P = Q(A), for some regular element a ∈ A we have ϕ(x) = ar(x) ∈ A[x]. But ϕ(α) = 0 and deg ϕ(x) < deg f(x) imply ϕ(x) = 0. So we obtain r(x) = 0 and g(x) ∈ f(x)P [x], as was be shown. (ii) Let P be a reduced ring. Since . \ A = A[x] (A[x] f(x)P [x]) =∼ (A[x] + f(x)P [x])f(x)P [x] ⊆ P, so we conclude that A is a reduced ring. Now by (i) we obtain that the ring A[α] is reduced. Conversely, suppose that A[α] is reduced. If P is not reduced and ϕ(x)+f(x)P [x] is its nontrivial nilpotent element, then we may assume that 0 6= ϕ(x) ∈ P [x], deg ϕ(x) < deg f(x) and ϕk(x) ∈ f(x)P [x] for some integer k > 1. Let a ∈ A be a nonzero regular element such that 0 6= aϕ(x) ∈ A[x]. Then it is clear that aϕ(x)+A[x] T f(x)P [x] is a nonzero nilpotent element of A. This shows that A is not reduced ring and by (i) 74 Reduced and irreducible simple algebraic extensions. . . we receive that A[α] is not reduced, which is a contradiction. So the proof is completed. Corollary 2. If the leading coefficient of the polynomial f(x) ∈ A[x] is an invertible element of A and f(x) is a minimal polynomial of α, then the rings A[α] and A[x]/f(x)A[x] are isomorphic. Proof. Let P be as above. Since the leading coefficient of the polynomial f(x) is invertible in A, it is easy to verify that f(x)A[x] ⊆ A[x] T f(x)P (x) ⊆ f(x)A[x]. Then the statement follows by Lemma 4(i).

3. Simple algebraic extensions of reduced rings Now let A be a reduced commutative ring and let f(x) ∈ A[x] be a minimal polynomial of the algebraic element α. It is clear that A[α] is reduced if and only if the ring B[α] is reduced for every finitely generated subring B ⊆ A such that f(x) ∈ B[x]. Therefore it is sufficient to find necessary and sufficient conditions A[α] to be reduce when A is a noetherian ring. First we shall consider the case when A is a field. Recall that a field F of characteristic p ≥ 0 is said to be perfect if p = 0, or p > 0 and F p = F ([2], p. 137). So every finite field and every algebraically closed field is perfect. If F is a field and f(x), g(x) ∈ F [x], then as ever, by (f, g) we shall denote the monic greatest common divisor over F of the polynomials f(x) and g(x). Moreover, f(x) and g(x) are associated if f(x) = ag(x) for some non zero element a ∈ F . Lemma 5. Let F be a field of characteristic p ≥ 0 and let f(x) be a nonzero polynomial over F . Then the following conditions are equivalent: (i) The quotient ring F [x]/f(x)F [x] is reduced. (ii) The polynomial f(x) is a product of distinct non associated irreducible polynomials over the field of F . (iii) Either (f, f 0) = 1, or F is not perfect field of characteristic p 6= 0 and (f, f 0) is a products of distinct non associated irreducible polynomials of the form ϕ(xp) ∈ F [x].

k1 k2 ks Proof. Let f(x) = af1 (x)f2 (x) ··· fs (x) be a factorization of f(x) over F , where f1(x), . . . , fs(x) are distinct non associated irreducible polynomials over F and a ∈ F . Since (fi, fj) = 1 for all i 6= j, by the Chinese theorem ([8], p.88) we have , s . ∼ X ki F [x] f(x)F [x] = ⊕F [x] fi (x)F [x]. i=1

It is clear that F [x]/f(x)F [x] is reduced if and only if k1 = k2 = ··· = ks = 1. So we obtain that (i) and (ii) are equivalent. S.V. Mihovski 75

Further, let f(x) = f1(x)f2(x) ··· fs(x) be a product of distinct non associated irreducible polynomials over F . Denote by g(x) = f1(x) ··· fk(x) the product of all factors of f(x), not having multiple roots. When k = 0 we put g(x) = 1. If k < s, let d(x) = fk+1(x)fk+2(x) ··· fs(x) be the product of all factors of f(x) which have multiple roots. This happens if p > 0 and p F is not perfect field (see ([2] p. 138). In such case fi(x) = ϕi(x ), where ϕi(x) ∈ F [x] for i = k + 1, k + 2, . . . , s. When k = s we put d(x) = 1. Thus f(x) = g(x)d(x) and either d(x) = 1, or d(x) = ϕ(xp) with ϕ(x) ∈ F [x] and deg ϕ(x) ≥ 1 (see [6] p. 162). Therefore we have d0(x) = 0. Since f 0(x) = g0(x)d(x) and (g, g0) = 1, it is clear that d(x) = (f, f 0). So we see that (ii) and (iii) are equivalent, as was to be shown. As an immediate consequence we obtain Corollary 3. Let F be a field and let f(x) ∈ F [x] be a nonzero polynomial. (i) If ∆(f) 6= 0, then the quotient ring F [x]/f(x)F [x] is reduced. (ii) If F is a perfect field, then the ring F [x]/f(x)F [x] is reduced if and only if ∆(f) 6= 0. Really, it is sufficient to observe that the conditions (f, f 0) = 1 and ∆(f) 6= 0 are equivalent. Now let A be a reduced commutative ring and let f(x) ∈ A[x] be a minimal polynomial of the algebraic element α. As was mentioned above, it is sufficient to find necessary and sufficient conditions under which A[α] is reduce, when A is a noetherian ring. Theorem 1. Let A be a reduced commutative noetherian ring with classical ring of quotients P = Q(A) and let α be an algebraic element over A with a minimal polynomial f(x) ∈ A[x]. If f(x) is a regular polynomial over A, then the following statements are equivalent: (i) The ring A[α] is reduced. (ii) For every regular element a ∈ A the polynomial af(x) is not divisible by squares of polynomials over A of degree t ≥ 1. (iii) For every minimal idempotent e ∈ P the polynomial ef(x) is a product of distinct non associated irreducible polynomials over the field eP . (iv) For every minimal idempotent e ∈ P , either (ef, ef 0) = e, or eP is a field of characteristic p > 0, eP is not a perfect field and (ef, ef 0) is a product of distinct non associated irreducible polynomials of the form ϕ(xp) ∈ eP [x]. Proof. Suppose that A ⊆ K and α ∈ K. By Lemma 3, without loss of generality, we may assume that P = Q(A) ⊆ K. Since A is a reduced commutative notherian ring, by Goldie’s Theorem (see [1], Corollary 2, p. 323), the ring P is a finite direct sum

P = A1 ⊕ A2 ⊕ · · · ⊕ Ak 76 Reduced and irreducible simple algebraic extensions. . . of ﬁelds Ai with identity elements ei (i = 1, . . . , k). Then Ai = eiP and it is clear that fi(x) = eif(x) is a minimal polynomial of α over the ﬁeld Ai for i = 1, . . . , k. Moreover,

A[α] ⊆ P [α] = A1[α] ⊕ A2[α] ⊕ · · · ⊕ Ak[α] and P [α] =∼ P [x]/f(x)P [x]. Thus, by Lemma 4(ii) we obtain that A[α] is reduced if and only if the rings A1[α],A2[α],...,Ak[α] are reduced. Hence, by Lemma 5 we conclude that the statements (i), (iii) and (iv) are equivalent. Therefore it is suﬃcient to prove that (i) and (ii) are equivalent. Really, suppose that A[α] is a reduced ring but af(x) = p2(x)q(x) for some regular element a ∈ A, where p(x), q(x) ∈ A[x] and deg p(x) ≥ 1. Then

p(x) = e1p(x) + e2p(x) + ··· + ekp(x) and without loss of generality we may assume that deg e1p(x) ≥ 1. Thus the equality af(x) = p2(x)q(x) shows that

2 e1af(x) = (e1p(x)) e1q(x) and

1 ≤ deg(e1p(x)q(x)) < deg(e1af(x)) = deg f1(x), where f1(x) = e1f(x). Therefore, e1p(x)q(x) + f1(x)A1[x] is a nonzero nilpotent element of the quotient ring A1[x]/f1(x)A1[x]. As far as f1(x) is a minimal polynomial of α over A1, by Corollary 2 we receive that A1[α] is not reduced, which is impossible. Conversely, if A[α] is not reduced ring, then P [α] is not reduced and without loss of generality we may assume that A1[α] is not reduced. Then by Corollary 2 and Lemma 5 we obtain that 2 f1(x) = p1(x)q1(x), where p1(x), q1(x) ∈ A1[x] and deg p1(x) ≥ 1. Now we put

p(x) = p1(x) + e2 + ··· + ek,

q(x) = q1(x) + f2 + ··· + fk and thus we receive f(x) = p2(x)q(x), where p(x), q(x) ∈ P [x] and deg p(x) ≥ 1. Since P is a ring of quotients, it follows that there exist regular elements b, c ∈ A such that bp(x) and cq(x) are elements of A[x]. Then a = b2c is a regular element in A and af(x) is divisible by the square of bp(x) ∈ A[x], as was to be showed. As was mentioned above, the main result of [11] asserts that if ∆(f) is a regular element in A, then A[α] is a reduced ring. But A[α] may be reduced even when ∆(f) = 0. Indeed, let α be a root of the polynomial f(x) = xp − y ∈ A[x], where A = F (y) is the ring of quotients of the polynomial ring F [y] over a field F of characteristic p > 0. Then ∆(f) = 0, S.V. Mihovski 77 f(x) is irreducible over A (see [2], p. 165) and A[α] is reduced by Corollary 2 and Lemma 5. We shall say that the reduced commutative ring A is locally perfect if for every finitely generated subring B ⊆ A and every minimal idempotent e ∈ Q(B) the field eQ(B) is perfect. If the additive group of the reduced ring A is either torsion free, or locally finite, then A is a locally perfect ring. Thus we have the following Corollary 4. Let α be an algebraic element over the commutative ring A with a regular minimal polynomial f(x) ∈ A[x] and let ∆(f) be the discriminant of f(x). (i) If ∆(f) is a regular element in A, then the ring A[α] is reduced if and only if A is reduced. (ii) If A is a reduced locally perfect ring, then A[α] is reduced if and only if ∆(f) is a regular element in A. Proof. (i) Assume that A is reduced and ∆(f) is regular in A, but A[α] is not reduced. If β is a nonzero nilpotent element of A[α], then let B be the subring of A, generated by the coefficients of β and f(x). Thus f(x) ∈ B[x] and β ∈ B[α]. Hence by the preceding theorem it follows that for some minimal idempotent e ∈ Q(B) the polynomial ef(x) has multiple roots and therefore ∆(ef) = 0. Since ∆(ef) = e∆(f), we obtain that the element ∆(f) is a proper divisor of zero, which is a contradiction. As far as the converse statement is trivial, the part (i) is proved. (ii) In view of (i) it is sufficient to prove that if A[α] is reduced, then ∆(f) is regular. Assume for moment that ∆(f)a = 0 and 0 6= a ∈ A. Let B be the finitely generated subring of A, generated by the coefficients of f(x) and the element a ∈ A. Thus f(x) ∈ B[x] and a ∈ B. Let e1, e2, . . . , en be a full orthogonal system minimal idempotents of Q(B). Then

∆(f) = ∆(e1f) + ∆(e2f) + ··· + ∆(enf), where ∆(eif) ∈ eiQ(B) for i = 1, 2, . . . , n. Since each eiQ(B) is a field and ∆(f) is a proper divisor of zero in B, we conclude that for some i (1 ≤ i ≤ n) we have ∆(eif) = 0. This implies that eif(x) has multiple roots. But eiQ(B) is a perfect field and by Corollary 2 and Lemma 5 we obtain that eiQ(B)[α] is not reduced and therefore Q(B)[α] is not reduced ring, which is a contradiction. So the proof is completed. Let K be any ring extension of the commutative ring A where K is not necessary commutative. If the element α ∈ K centralizes A, that is α.a = a.α for all a ∈ A, then we may to consider the simple commutative ring extension A[α]. So we have the following Corollary 5. Let F be a perfect field and let S be an element of the n × n matrix ring M(n, F ). If F contains all characteristic values of S, then the 78 Reduced and irreducible simple algebraic extensions. . . ring F [S] is reduced if and only if for some non-singular matrix T ∈ M(n, F ) the matrix TST −1 is diagonal. Proof. By Corollary 2 and Corollary 3(ii) the ring F [S] is reduced if and only if ∆(f) 6= 0 where f = f(λ) is the minimal polynomial of S in F [λ]. Since f(λ) is the last invariant factor of the characteristic matrix S − λE (see [5], p. 389), this condition is equivalent with the condition the Jordan’s normal form of S to be diagonal.

4. Simple algebraic extensions of irreducible commutative rings In this part we shall study the problem who the ring A[α] contains nontrivial idempotent elements. Later on we shall say that the idempotent E of the ring A[α] (respectively of A[x]/f(x)A[x]) is a trivial idempotent if E is an element of the subring A (respectively of the subring (A+f(x)A[x])/f(x)A[x]). As usually we shall say that the polynomial p(x) ∈ A[x] divides the polynomial f(x) ∈ A[x] over the ring A if there exists a polynomial q(x) ∈ A[x] such that f(x) = p(x)q(x). The polynomial p(x) ∈ A[x] is said to be a trivial divisor of f(x) if p(x) divides f(x) and there exists an element a ∈ A such that p(x) + f(x)A[x] = a + f(x)A[x], that is f(x) divides p(x) − a over A. For example, if e ∈ A is a nontrivial idempotent of A, then every polynomial f(x) ∈ A[x] has a trivial decomposition

f(x) = [ef(x) + (1 − e)][e + (1 − e)f(x)]. The decomposition f(x) = p(x)q(x) over A is said to be nontrivial decomposition if over A the polynomials p(x) and q(x) are nontrivial divisors of f(x). Also, the decomposition f(x) = p(x)q(x) is an essential decomposition over A if p(x) and q(x) are nontrivial divisors of f(x) and deg p(x) < deg f(x), deg q(x) < deg f(x). We shall say that the polynomial f(x) is irreducible over the ring A if f(x) has no nontrivial decomposition over A. Recall that if F is a ﬁeld and ϕ(x), ψ(x) ∈ F [x], then for the greatest common divisor (ϕ, ψ) there exist polynomials u(x), v(x) ∈ F [x] such that (ϕ, ψ) = u(x)ϕ(x) + v(x)ψ(x) and deg u(x) < deg ψ(x), deg v(x) < deg ϕ(x). Likewise, if A is any commutative ring and ϕ(x), ψ(x) ∈ A[x], then by Lemma 1 it follows that for the resultant R(ϕ, ψ) there exist polynomials u(x), v(x) ∈ A[x] such that

R(ϕ, ψ) = u(x)ϕ(x) + v(x)ψ(x) ∈ A and deg u(x) < deg ψ(x), deg v(x) < deg ϕ(x). From here on we shall use these facts without special stipulations. S.V. Mihovski 79

Let f(x) be a minimal polynomial over the ﬁeld F of the algebraic element α. Since the rings F [α] and F [x]/f(x)F [x] are isomorphic, by Chain’s theorem it follows that F [α] contains nontrivial idempotents if and only if f(x) is not associated with a power of some irreducible polynomial over F . Now we shall prove the following lemma, which gives the idempotents of F [α] in explicit form. Lemma 6. Let α be an algebraic element over the ﬁeld F with a minimal polynomial f(x) ∈ F [x]. Then (i) The ring F [α] is irreducible if and only if f(x) is associated with a power of an irreducible polynomial of F [x]. (ii) The elements E1(α) and E2(α) of F [α] form a full orthogonal system idempotents if and only if over F there exists a decomposition f(x) = ϕ(x)ψ(x) such that

(ϕ, ψ) = u(x)ϕ(x) + v(x)ψ(x) = 1, where deg(u(x)ϕ(x)) < deg f(x) and

E1(α) = u(α)ϕ(α),E2(α) = v(α)ψ(α).

(iii) The elements E1(α) and E2(α) of F [α] form a nontrivial full orthogonal system idempotents if and only if over F there exists an essential decomposition f(x) = ϕ(x)ψ(x) such that

R(ϕ, ψ) = u1(x)ϕ(x) + v1(x)ψ(x) 6= 0 and −1 −1 E1(α) = R(ϕ, ψ) u1(α)ϕ(α),E2(α) = R(ϕ, ψ) v1(α)ψ(α). Proof. (i) Let F [α] be an irreducible ring and let

k1 k2 ks f(x) = ap1 (x)p2 (x) ··· ps (x) be the canonical decomposition of f(x) over the ﬁeld F . Since F [α] and F [x]/f(x)F [x] are isomorphic rings, by the Chinese theorem we obtain that k1 f(x) = ap1 (x) and therefore f(x) is associated with a power of irreducible polynomial over F . Conversely, if f(x) is associated with a power of an irreducible polynomial over F and f(x) = apk(x), then f(x)F [x] = pk(x)F [x] and p(x)F [x]pk(x)F [x] is a nilideal of F [x]pk(x)F [x]. Since

. . . (F [x] pk(x)F [x]) (p(x)F [x] pk(x)F [x]) =∼ F [x]/p(x)F [x] and F [x]/p(x)F [x] is a ﬁeld, by [4], Proposition 11.5.1 we conclude that (i) follows. 80 Reduced and irreducible simple algebraic extensions. . .

(ii)Suppose that the elements E1(α) and E2(α) form a full orthogonal system idempotents of F [α]. Since the rings F [α] and F¯[x] = F [x]f(x)F [x] are isomorphic, it follows that in F¯[x] there exist elements

E¯1(x) = e1(x) + f(x)F [x], E¯2(x) = e2(x) + f(x)F [x]

such that E¯1(x) and E¯2(x) form a full orthogonal system idempotents of F¯[x] and e1(α) = E1(α), e2(α) = E2(α). Without loss of generality we may to assume that deg ei(x) < deg f(x) for i = 1, 2. Obviously, E¯1(x) and E¯2(x) form a full orthogonal system idempotents if and only if e1(x) + e2(x) = 1 and e1(x)e2(x) = f(x)q(x) for some q(x) ∈ F [x]. If E¯1(x) and E¯2(x) form a trivial system orthogonal idempotents of F¯[x] and e1(x) = 0, e2(x) = 1, then we put ϕ(x) = f(x), ψ(x) = 1 and u(x) = 0, v(x) = 1. Suppose that E¯1(x) and E¯2(x) form a nontrivial system orthogonal idempotents. Since F [x] is a factorial ring (see [8], p. 142), we conclude that

f(x) = ϕ(x)ψ(x), e1(x) = u(x)ϕ(x), e2(x) = v(x)ψ(x), where ϕ(x) = (e1, f) and ψ(x) = (e2, f). Moreover, the polynomials u(x) and v(x) in F [x] are uniquely determined. Since the converse statement is trivial, so (ii) is proved. (iii) When E¯1(x) and E¯2(x) form a nontrivial system orthogonal idempotents of F¯[x], it is clear that 0 < deg ei(x) < deg f(x) for i = 1, 2. Thus we obtain that the decomposition f(x) = ϕ(x)ψ(x) is nontrivial and therefore deg ϕ(x) ≥ 1, deg ψ(x) ≥ 1. As far e1(x) + e2(x) = 1, we have (ϕ, ψ) = 1 and hence we receive

R(ϕ, ψ) = u1(x)ϕ(x) + v1(x)ψ(x) 6= 0, where, by Lemma 1, deg u1(x) < deg ψ(x) and deg v1(x) < deg ϕ(x). Now −1 −1 it is easy to verify that u(x) = R(ϕ, ψ) u1(x) and v(x) = R(ϕ, ψ) v1(x). So we prove and the statement (iii). The following lemma is an analog of the parts (ii) and (iii) of the preceding lemma for commutative artinian rings. Lemma 7. Let α be an algebraic element over the reduced commutative artinian ring A with a regular minimal polynomial f(x) ∈ A[x]. Then (i) The element E(α) is a nontrivial idempotent in A[α] if and only if over A there exists a nontrivial decomposition f(x) = ϕ(x)ψ(x) such that u(x)ϕ(x) + v(x)ψ(x) = 1 for some polynomials u(x) and v(x) of A[x], where deg(u(x)ϕ(x)) < deg f(x) and E(α) = u(α)ϕ(α). (ii) The ring A[α] contains nontrivial idempotents if and only if for some nonzero idempotent e ∈ A over the ring eA there exists an essential decomposition ef(x) = ϕ(x)ψ(x) such that R(ϕ, ψ) is a nonzero element of eA. S.V. Mihovski 81

Proof. By Wedderburn-Artin theorem, A = F1 ⊕ F2 ⊕ · · · ⊕ Fm is a ﬁnite direct sum of ﬁelds Fi with identity elements ei (i = 1, . . . , m). So we have the decomposition

A[α] = F1[α] ⊕ F2[α] ⊕ · · · ⊕ Fm[α]. (i) Suppose that E(α) is a nontrivial idempotent of A[α]. Then the elements E1(α) = E(α) and E2(α) = 1 − E(α) have the decompositions

(6) Ek(α) = Ek1(α) + Ek2(α) + ··· + Ekm(α)(k = 1, 2), where E1i(α) and E2i(α) form a full orthogonal system idempotents of Fi[α]. Obviously fi(x) = eif(x) is a minimal polynomial of α over the ﬁeld Fi = eiA for all i = 1, 2, . . . , m. By Lemma 6(ii) it follows that over Fi there exists a decomposition eif(x) = ϕi(x)ψi(x) such that

(ϕi, ψi) = ui(x)ϕi(x) + vi(x)ψi(x) = ei, where deg(ui(x)ϕi(x)) < deg f(x) and

E1i(α) = ui(α)ϕi(α),E2i(α) = vi(α)ψi(α) for i = 1, 2, . . . , m. Then f(x) = ϕ(x)ψ(x), where

ϕ(x) = ϕ1(x) + ϕ2(x) + ··· + ϕm(x),

ψ(x) = ψ1(x) + ψ2(x) + ··· + ψm(x). Moreover, u(x)ϕ(x) + v(x)ψ(x) = 1, where

u(x) = u1(x) + u2(x) + ··· + um(x),

v(x) = v1(x) + v2(x) + ··· + vm(x). Obviously, deg(u(x)ϕ(x)) < deg f(x) and E(α) = u(α)ϕ(α). Since the converse statement is evident, so (i) is proved. (ii) If E(α) is a nontrivial idempotent of A[α], then again we put E1(α) = E(α) and E2(α) = 1 − E(α). Suppose that E1(α) and E2(α) have the decompositions (6). Without loss of generality we may to assume that E11(α) and E21(α) form a full nontrivial orthogonal system idempotents of F1[α], where f1(x) = ef(x) is a minimal polynomial of α over the ﬁeld F1 = eA and e = e1. Then by Lemma 6(ii), over F1 there exists an essential decomposition ef = ϕ(x)ψ(x) such that R(ϕ, ψ) 6= 0. Conversely, if for some idempotent e ∈ A over the ring eA there exists an essential decomposition ef(x) = ϕ(x)ψ(x) such that R(ϕ, ψ) 6= 0, we shall have the decompositions

ϕ(x) = ϕ1(x) + ϕ2(x) + ··· + ϕm(x),

ψ(x) = ψ1(x) + ψ2(x) + ··· + ψm(x), where ϕi(x) = eiϕ(x) and ψi(x) = eiψ(x) for i = 1, . . . , m. Since 82 Reduced and irreducible simple algebraic extensions. . .

R(ϕ, ψ) = R(ϕ1, ψ1) + R(ϕ2, ψ2) + ··· + R(ϕm, ψm) 6= 0, it follows that for some k (1 ≤ k ≤ m) we have R(ϕk, ψk) 6= 0. Then

ekef(x) = ekf(x) = ϕk(x)ψk(x) is an essential decomposition over the field Fk and by Lemma 6(ii) it follows that Fk[α] contains nontrivial idempotents. Since Fk[α] ⊆ A[α], we conclude that A[α] contains nontrivial idempotents, as was to be shoved. Let I be an ideal of A and let I[α] be the simple algebraic extension of I, which is obtained by adjoining of α to I. As for A[x]/f(x)A[x], we shall say that an idempotent E of the ring A[α]/I[α] is trivial if E is an element of the subring (A + I[α])/I[α]. Lemma 8. Let A[α] be any simple ring extension of the commutative ring A and let I be a nil-ideal of A. (i) All idempotents of A[α] are trivial if and only if all idempotents of the quotient ring A[α]/I[α] are trivial. (ii) If α is an algebraic element over the ring A with a regular minimal polynomial n n−1 f (x) = a0x + a1x + ··· + an−1x + an (n ≥ 1), then there exists a simple algebraic extension A¯[β] of the quotient ring A¯ = A/I, such that A[α]I[α] =∼ A¯[β] and ¯ n n−1 f(x) =a ¯0x +a ¯1x + ··· +a ¯n−1x +a ¯n (¯ak = ak + I) is a regular minimal polynomial over A¯ of the element β. Proof. (i) Suppose that all idempotents of A[α] are trivial and let E(α) = u(α) + I[α] be an idempotent of A[α]/I[α]. Then u(α)2 − u(α) ∈ I[α] and, since I[α] is a nil-ideal of A[α], by ([4], Proposition 11.5.1) it follows that there exists an idempotent e(α) ∈ A[α] such that e(α) − u(α) ∈ I[α]. Therefore E(α) = e(α) + I[α] and, since all idempotents of A[α] are trivial, we have e(α) = e ∈ A. So we conclude that all idempotents of A[α]/I[α] are trivial. Conversely, assume that all idempotents of A[α]/I[α] are trivial. If e(α) is an idempotent of A[α], then e(α) + I[α]is a trivial idempotent of A[α]/I[α] and hence for some element a ∈ A we have e(α) + I[α] = a + I[α]. Since e(α)2 − e(α) = 0, we obtain that a2 − a ∈ I. Then again by ([4], Proposition 11.5.1) we obtain that there exists an idempotent e ∈ A, such that a − e ∈ I. As far I ⊆ I[α], we receive e(α) + I[α] = e + I[α]. Suppose that e(α) = e+v(α), where v(α) ∈ I[α] is a nilpotent element. Then e(α)e = e + v(α)e is an invertible element of the ring eA[α]. But e(α)e is simultaneously an idempotent of eA[α]. Thus we obtain that e(α)e = e and v(α)e = 0. Now e(α)(1 − e) = v(α)(1 − e) is simultaneously an idempotent S.V. Mihovski 83 and a nilpotent element of (1 − e)A[α]. So we conclude that v(α)(1 − e) = 0 and hence v(α) = v(α)e + v(α)(1 − e) = 0. Therefore e(α) = e is a trivial idempotent of A[α] and thus (i) is proved. . (ii) Obviously, A[α] I[α] = A˜[˜α] is a simple ring extension of the subring A˜ = (A + I[α])/I[α], obtained by adjoining of the element α˜ = α + I[α] to A˜ . Since f(α) = 0, it is clear that α˜ is a root of ˜ n n−1 f(x) =a ˜0x +a ˜1x + ··· +a ñ−1x +a ñ (ãk = ak + I[α]). Therefore A˜[˜α] is a simple algebraic extension of A˜. If

˜ m ˜ m−1 ˜ ˜ ˜ g˜(x) = b0x + b1x + ··· + bm−1x + bm (bk = bk + I[α] ∈ A˜) is a minimal nonzero polynomial of α˜ over A˜, then b0 ∈/ I and m ≤ n. Suppose that m < n. As far g˜(˜α) = g(α) + I[α] = I[α], we conclude that

m m−1 g(α) = b0α + b1α + ··· + bm−1α + bm s s−1 = c0α + c1α + ··· + cs−1α + cs (ck ∈ I), where b0 6= c0. Now we use the fact that a0 is an invertible element in the ring of quotients Q(A) and f(α) = 0. So without loss of generality we may to suppose that s < n and c0, c1, . . . , cs ∈ NilQ(A). Therefore there exists a regular element a ∈ A such that abi ∈ A (i = 1, . . . , m) and acj ∈ I (j = 1, . . . , s). Thus we obtain that α is a root of a nonzero polynomial of degree t = min{m, s} < n, which is impossible. Hence m = n and f˜(x) is a minimal polynomial of α˜ overA˜. By a similar way we prove that T A I[α] = I. Then it is easy to verify that a˜0 is a regular element of A˜. Moreover, . \ A˜ = (A + I[α])/I[α] =∼ A (A I[α]) = AI = A.¯ Let f¯(x) be a minimal polynomial of some element β over the ring A¯. Then the mapping A˜[˜α] → A¯[β], defined by α˜ 7→ β and a + I[α] 7→ a + I for a ∈ A is an isomorphism, as was to be showed. Now by Lemma 8(ii) we shall prove following theorem. Theorem 2. Let α be an algebraic element over an artinian commutative ring A with a regular minimal polynomial f(x) ∈ A[x]. If A¯ = A/NilA and f¯(x) is the natural image of f(x) into A¯[x], then (i) A[α] is irreducible if and only if A¯ is a field and f¯(x) is associated with a power of some irreducible polynomial over A¯. (ii) A[α] contains only trivial idempotents if and only if for every minimal idempotent e¯ ∈ A¯ the polynomial e¯f¯(x) is associated with a power of some irreducible polynomial over the field e¯A¯. 84 Reduced and irreducible simple algebraic extensions. . .

Proof. If A is an artinian commutative ring, then A¯ = A/NilA is a finite direct sum of fields [6, 7]. Since A is irreducible if and only if A¯ is irreducible, by Lemma 8 we obtain that A[α] is irreducible if and only if A¯[β] is irreducible, where A¯ is a field and f¯(x) is a regular minimal polynomial of β over the field A¯. Then the statement (i) follows by Lemma 7(i). Again by Lemma 8 it follows that A[α] contains only trivial idempotents if and only if A¯[β] contains only trivial idempotents. Since A¯ is a finite direct sum of fields, by Lemma 7(i) we conclude that for every minimal idempotent e¯ ∈ A¯ the ring e¯A¯[β] contains only trivial idempotents. So by Lemma 8 we obtain and the statement (ii). Theorem 3. Let α be an algebraic element over a commutative noetherian ring A with a monic minimal polynomial f(x) ∈ A[x] and let P = Q(A¯) be the ring of quotients of A¯ = A/NilA. The ring A[α] contains nontrivial idempotents if and only if over the ring P¯ there exists a nontrivial decomposition f¯(x) =ϕ ¯(x)ψ¯(x) such that u¯(x)ϕ ¯(x) +v ¯(x)ψ¯(x) = 1¯ for some polynomials u¯(x), v¯(x) ∈ P [x], where deg(¯u(x)ϕ ¯(x)) < deg f¯(x) and u¯(x)ϕ ¯(x) ∈ A¯[x].

Proof. Suppose that A[α] contains a nontrivial idempotent E(α). Then by Lemma 8(i) it follows that E(α) + I[α] is a nontrivial idempotent of A[α]/I[α], where I = NilA. Now by Lemma 8(ii) we conclude that there exists a nontrivial idempotent E¯(β) of the ring A¯[β], where f¯(x) ∈ A¯[x] is a minimal polynomial of β. Without loss of generality, by Lemma 3 we may assume that E¯(β) is a nontrivial idempotent of P [β]. Since P is a reduced artinian ring, by Lemma 7(i) we conclude that over the ring P there exists a nontrivial decomposition f¯(x) =ϕ ¯(x)ψ¯(x) such that u¯(x)ϕ ¯(x)+v ¯(x)ψ¯(x) = 1¯ for some polynomials u¯(x) and v¯(x) of P [x], where deg(¯u(x)ϕ ¯(x)) < deg f¯(x) and E¯(β) =u ¯(β)ϕ ¯(β) ∈ A¯[β]. Since f¯(x) ∈ A¯[x] is a monic minimal polynomial of β over A¯ and deg(¯u(x)ϕ ¯(x)) < deg f¯(x), it is clear that E¯(β) =u ¯(β)ϕ ¯(β) ∈ A¯[β] implies u¯(x)ϕ ¯(x) ∈ A¯[x]. Conversely, suppose that the polynomial f(x) ∈ A[x] satisfy the conditions of the theorem and let g(x) be a polynomial in A[x] such that g¯(x) =u ¯(x)ϕ ¯(x). If β is an algebraic element over A¯ with a minimal polynomial f¯(x) ∈ B¯[x], then by Lemma 8(ii) it follows that A[α]/I[α] and A¯[β] are isomorphic rings. Since g¯(β) =u ¯(β)ϕ ¯(β) is a nontrivial idempotent in A¯[β], the element g(α) + I[α] is a nontrivial idempotent in A[α]/I[α] (see the proof of Lemma 8(ii)). If u = g2(α) − g(α), then by Proposition 3.6.1 [7] we conclude that E(α) = g(α) − x[1 − 2g(α)] is a nontrivial idempotent of A[α], where

1 4 6 x = 2u − u2 + u3 − · · · . 2 2 3

So the theorem is proved. S.V. Mihovski 85

It is easy to verify that in the preceding theorem the condition f(x) to be a monic polynomial is not necessary. Really, let f(x) = 4x2 − 1 be a minimal polynomial of the algebraic element α over the integer ring Z. Then f(x) = (2x−1)(2x+1) is a nontrivial decomposition over the ﬁeld Q = Q(Z) and 2−1(2x + 1) − 2−1(2x − 1) = 1. Thus e(x) = 2−1(2x + 1) = x + 2−1 is not element of Z[x], but e(α) = α + 2−1 = α + 2α2 is an idempotent of Z[α]. For regular minimal polynomials we have the following Corollary 6. Let A be a commutative noetherian ring and let P = Q(A¯) be the ring of quotients of A¯ = A/NilA. Suppose that f(x) is a regular minimal polynomial of an algebraic element α over the ring A. If for every minimal idempotent e ∈ P the polynomial ef(x) is associated with a power of some irreducible polynomial over the ﬁeld eP , then all idempotents of the ring A[α] are trivial. The proof of this corollary is as the proof of Theorem 3.

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S.V. Mihovski University of Plovdiv “P.Hilendarski” Tsar Assen str. 24 Plovdiv 4000 Bulgaria E-mail address: [email protected]